\newproblem{lay:3_2_14}{
  % Problem identification
	\begin{large}
	  \hspace{\fill}\newline
    \textbf{Lay, 3.2.14}
	\end{large}
	\\
  \ifthenelse{\boolean{identifyAuthor}}{\textit{Carlos Oscar Sorzano, Aug. 31st, 2013} \\}{}

  % Problem statement
  Combine the methods of row reduction and cofactor expansion to compute the determinant 
	$\left|\begin{array}{rrrr}-3 & -2 & 1 &-4 \\ 1 & 3 & 0 & -3\\ -3 & 4 & -2 & 8 \\ 3 & -4 & 0 & 4\end{array}\right|$
}{
   % Solution
	\begin{center}
		\begin{tabular}{cl}
			& $D=\left|\begin{array}{rrrr}-3 & -2 & 1 &-4 \\ 1 & 3 & 0 & -3\\ -3 & 4 & -2 & 8 \\ 3 & -4 & 0 & 4\end{array}\right|$ \\
			$\mathbf{r}_3\leftarrow \mathbf{r}_3+2\mathbf{r}_1$ & $D=\left|\begin{array}{rrrr}-3 & -2 & 1 &-4 \\ 1 & 3 & 0 & -3\\ -9 & 0 & 0 & 0 \\ 3 & -4 & 0 & 4\end{array}\right|
			  = (-1)^{1+3}1\left|\begin{array}{rrr} 1 & 3 & -3\\ -9 & 0 & 0 \\ 3 & -4 & 4\end{array}\right|$ \\
			& $D=\left|\begin{array}{rrr} 1 & 3 & -3\\ -9 & 0 & 0 \\ 3 & -4 & 4\end{array}\right|=(-1)^{2+1}(-9)\left|\begin{array}{rrr} 3 & -3\\ -4 & 4\end{array}\right|$ \\
			& $D=9(3\cdot 4 - (-3)\cdot (-4))=0$ \\
		\end{tabular}
	\end{center}
}
\useproblem{lay:3_2_14}
\ifthenelse{\boolean{eachProblemInOnePage}}{\newpage}{}
